Reflecting About A Line- Solids

[Justabeginner]

Homework Statement

Find the volume of the solid generated by revolving the region bounded by the graphs of $y= x^2 - 4x + 5$ and $y= 5- x$ about the line $y= -1$

Homework Equations

That is what I'm trying to figure out here, which one of the methods to use.

The Attempt at a Solution

I have these steps so far...

x^2 - 4x + 5= 5 - x
x^2 - 3x = 0
x(x-3)= 0
x= 0 and x= 3
These are my intersection points. From 0 to 3 on this graph, the function 5- x has greater y-values than those of x^2- 4x + 5, so I get (5-x)- (x^2-4x+5)= 3x-x^2. And now I am stumped. What method would I use? And how am I supposed to know if it's a disk, shell, or washer method looking at the graph? I am clueless on how to do that. Thank you for your help.

 

[CAF123]

Generally it will be easier to do the question via the method of shells or via disks/washers. Boths methods work. I think using disks is easier in this case,

 

[Mark44]

Homework Statement

Find the volume of the solid generated by revolving the region bounded by the graphs of $y= x^2 - 4x + 5$ and $y= 5- x$ about the line $y= -1$

Homework Equations

That is what I'm trying to figure out here, which one of the methods to use.

The Attempt at a Solution

I have these steps so far...

x^2 - 4x + 5= 5 - x
x^2 - 3x = 0
x(x-3)= 0
x= 0 and x= 3
These are my intersection points. From 0 to 3 on this graph, the function 5- x has greater y-values than those of x^2- 4x + 5, so I get (5-x)- (x^2-4x+5)= 3x-x^2. And now I am stumped. What method would I use? And how am I supposed to know if it's a disk, shell, or washer method looking at the graph? I am clueless on how to do that. Thank you for your help.

You can always use either method (shells vs. disks/washers). The question really is which of the two methods results in an integral that is easier to evaluate.

I find it helpful in these problems to draw two graphs. The first shows the region that is being revolved; the second shows a cross-section of the solid of revolution, with some detail on the typical volume element (a shell or disk/washer), including its dimensions.

 

[Justabeginner]

So using the disk method which if I'm not mistaken is (f(x)^2) dx, I should get:

(3x-x^2)^2= 9x^2 - 6x^3 + x^4 with a and b being 0 and 3 respectively.

108- [0]= 108?

But when I draw the graph.. I don't see how the y= -1 would affect this at all then?

 

[LCKurtz]

So using the disk method which if I'm not mistaken is (f(x)^2) dx, I should get:

(3x-x^2)^2= 9x^2 - 6x^3 + x^4 with a and b being 0 and 3 respectively.

108- [0]= 108?

But when I draw the graph.. I don't see how the y= -1 would affect this at all then?

Which means it's wrong, eh? The formula$$
\int_a^b \pi f^2(x)dx$$is for the area under $y=f(x)$ rotated about the $x$ axis. That isn't what you have, and you forgot the $\pi$ anyway. You want the formula$$
V=\int_a^b \pi (r_{upper}^2 - r_{lower}^2)\, dx$$where $r$ is the radius of revolution for the appropriate curve. The axis being $y=-1$ will affect that.

 

[Justabeginner]

So by using the correct formula, I get R = (5-x)+1 and r = (x^2-4x+5)+1

Then after I plug in and solve, V= 162pi/5. Is this correct?

 

[Mark44]

It would be helpful if you showed the work you did to get that result. I'm not saying that it's wrong, but I can't tell without duplicating your effort.

 

[Justabeginner]

Well I did this:

pi * integral { (5-x)+1)^2) with a=0 and b=2 - ((x^2-4x+6)^2)} dx

Then I just simplified.

 

[Mark44]

This looks good so far.

Here's your integral in LaTeX.
$$ \pi \int_0^3 [(6 - x)^2 - (x^2 - 4x + 6)^2]dx$$

There are lots of opportunities for algebra errors to creep in, so show us the rest of your work.

Click the Quote button to see what I did